__Question 7.8__:A non-uniform bar of weight

*W*is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance

*d*of the centre of gravity of the bar from its left end.

__Solution__:The free body diagram of the bar is shown in the following figure.

Length of the bar,

*l*= 2 m

*T*

_{1 }and

*T*

_{2}are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have:

T

_{1}Sin 36.9

^{0}= T

_{2}Sin 53.1

^{0}

T

_{1}/ T

_{2}= 4 / 3

⇒ T

_{1}= (4/3) T

_{2}

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

T

_{1}(Cos 36.9) X d = T

_{2}Cos 53.1 (2 - d)

T

_{1}X 0.800 d = T

_{2}X 0.600 (2 - d)

(4/3) X T

_{2}X 0.800d = T

_{2}(0.600 X 2 - 0.600d)

1.067d + 0.6d = 1.2

∴ d = 1.2 / 1.67

= 0.72 m

Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.

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